Show this inequality $fracna_1 - a_0 + fracn - 1a_2 - a_1 + cdots + frac1a_n - a_n-1 ge sum_k=1^n frack^2a_k$How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Prove that: $sqrt[3]a_1^3+ a_2^3 +cdots+a_n^3 le sqrta_1^2 + a_2^2 +cdots+a_n^2$Let $a_1,a_2,a_3,…a_ngeq0$ and $a_1+a_2+a_3+…+a_n=2m$,then prove that the greatest possible value of $sum_i=1^na_i.a_i+1$.Prove that $tfraca_1a_2cdots a_n(1-a_1-a_2-cdots-a_n)(a_1+a_2+cdots+a_n)(1-a_1)cdots(1-a_n) leq frac1n^n+1.$Prove that $fraca_1^2a_1+b_1+cdots+fraca_n^2a_n+b_n geq frac12(a_1+cdots+a_n).$Proving the inequality $frac11+a_1 + frac2(1+a_1)(1+a_2) + cdots + fracn(1+a_1)ldots (1+a_n) < 2$$a_1 + a_2 + dots a_n = 1$ find min of $a_1^2 +fraca_2^22 + dots + fraca_n^2n.$Prove that $a_1!cdots a_k! < n!$ whenever $a_1+cdots+a_k < n$Inequality problem: $(a_1+a_2+dots+a_n)^2 le n^2 (a_1^2+a_2^2+dots+a_n^2)$How can I prove that $(a_1+a_2+dotsb+a_n)(frac1a_1+frac1a_2+dotsb+frac1a_n)geq n^2$
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Show this inequality $fracna_1 - a_0 + fracn - 1a_2 - a_1 + cdots + frac1a_n - a_n-1 ge sum_k=1^n frack^2a_k$
How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Prove that: $sqrt[3]a_1^3+ a_2^3 +cdots+a_n^3 le sqrta_1^2 + a_2^2 +cdots+a_n^2$Let $a_1,a_2,a_3,…a_ngeq0$ and $a_1+a_2+a_3+…+a_n=2m$,then prove that the greatest possible value of $sum_i=1^na_i.a_i+1$.Prove that $tfraca_1a_2cdots a_n(1-a_1-a_2-cdots-a_n)(a_1+a_2+cdots+a_n)(1-a_1)cdots(1-a_n) leq frac1n^n+1.$Prove that $fraca_1^2a_1+b_1+cdots+fraca_n^2a_n+b_n geq frac12(a_1+cdots+a_n).$Proving the inequality $frac11+a_1 + frac2(1+a_1)(1+a_2) + cdots + fracn(1+a_1)ldots (1+a_n) < 2$$a_1 + a_2 + dots a_n = 1$ find min of $a_1^2 +fraca_2^22 + dots + fraca_n^2n.$Prove that $a_1!cdots a_k! < n!$ whenever $a_1+cdots+a_k < n$Inequality problem: $(a_1+a_2+dots+a_n)^2 le n^2 (a_1^2+a_2^2+dots+a_n^2)$How can I prove that $(a_1+a_2+dotsb+a_n)(frac1a_1+frac1a_2+dotsb+frac1a_n)geq n^2$
$begingroup$
For $a_1, ldots , a_n in mathbbR, a_1 < a_2 < cdots <a_n$ and $a_i ne 0$, show that
$dfracna_1 - a_0 + dfracn - 1a_2 - a_1 + cdots + dfrac1a_n - a_n-1 ge sum_k=1^n dfrack^2a_k$
where $a_0 = 0$.
I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.
The inequality mentioned in the chapters are
Cauchy-Schwarz Inequality
Weierstrass's Inequality
Tchebychev's Inequality
I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.
inequality cauchy-schwarz-inequality geometric-inequalities
edited Mar 21 at 17:40
StubbornAtom
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
$endgroup$
add a comment |
$begingroup$
For $a_1, ldots , a_n in mathbbR, a_1 < a_2 < cdots <a_n$ and $a_i ne 0$, show that
$dfracna_1 - a_0 + dfracn - 1a_2 - a_1 + cdots + dfrac1a_n - a_n-1 ge sum_k=1^n dfrack^2a_k$
where $a_0 = 0$.
I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.
The inequality mentioned in the chapters are
Cauchy-Schwarz Inequality
Weierstrass's Inequality
Tchebychev's Inequality
I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.
inequality cauchy-schwarz-inequality geometric-inequalities
edited Mar 21 at 17:40
StubbornAtom
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
$endgroup$
1
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
1
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51
add a comment |
$begingroup$
For $a_1, ldots , a_n in mathbbR, a_1 < a_2 < cdots <a_n$ and $a_i ne 0$, show that
$dfracna_1 - a_0 + dfracn - 1a_2 - a_1 + cdots + dfrac1a_n - a_n-1 ge sum_k=1^n dfrack^2a_k$
where $a_0 = 0$.
I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.
The inequality mentioned in the chapters are
Cauchy-Schwarz Inequality
Weierstrass's Inequality
Tchebychev's Inequality
I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.
inequality cauchy-schwarz-inequality geometric-inequalities
edited Mar 21 at 17:40
StubbornAtom
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
$endgroup$
For $a_1, ldots , a_n in mathbbR, a_1 < a_2 < cdots <a_n$ and $a_i ne 0$, show that
$dfracna_1 - a_0 + dfracn - 1a_2 - a_1 + cdots + dfrac1a_n - a_n-1 ge sum_k=1^n dfrack^2a_k$
where $a_0 = 0$.
I tried mathematical induction but not able to solve (not able to simplify n = k +1) expression.
The inequality mentioned in the chapters are
Cauchy-Schwarz Inequality
Weierstrass's Inequality
Tchebychev's Inequality
I think we need to use Tchebychev's Inequality to prove this but I'm not able to solve this.
inequality cauchy-schwarz-inequality geometric-inequalities
inequality cauchy-schwarz-inequality geometric-inequalities
edited Mar 21 at 17:40
StubbornAtom
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
edited Mar 21 at 17:40
StubbornAtom
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
edited Mar 21 at 17:40
StubbornAtom
6,29831440
edited Mar 21 at 17:40
StubbornAtom
6,29831440
edited Mar 21 at 17:40
StubbornAtom
6,29831440
6,29831440
asked Mar 21 at 16:57
JekiJeki
53
asked Mar 21 at 16:57
JekiJeki
53
asked Mar 21 at 16:57
JekiJeki
53
53
1
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
1
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51
add a comment |
1
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
1
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51
1
1
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
1
1
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We proceed by induction on $n$. For $n = 1$ we have
$$
frac1a_1 - a_0 geq frac1a_1,
$$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
beginalign*
sum_k = 1^n + 1fracn + 2 - ka_k - a_k - 1 &= sum_k = 1^nfracn + 1 - ka_k - a_k - 1 + sum_k = 1^n + 1frac1a_k - a_k - 1 \
& geq sum_k = 1^n frack^2a_k + sum_k = 1^n + 1frac1a_k - a_k - 1,
endalign*
by the induction hypothesis. It is therefore sufficient to prove that
$$
sum_k = 1^n + 1frac1a_k - a_k - 1 geq frac(n + 1)^2a_n + 1.
$$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, ldots, a_n - a_n - 1$.
answered Mar 21 at 20:42
P. SendenP. Senden
751212
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We proceed by induction on $n$. For $n = 1$ we have
$$
frac1a_1 - a_0 geq frac1a_1,
$$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
beginalign*
sum_k = 1^n + 1fracn + 2 - ka_k - a_k - 1 &= sum_k = 1^nfracn + 1 - ka_k - a_k - 1 + sum_k = 1^n + 1frac1a_k - a_k - 1 \
& geq sum_k = 1^n frack^2a_k + sum_k = 1^n + 1frac1a_k - a_k - 1,
endalign*
by the induction hypothesis. It is therefore sufficient to prove that
$$
sum_k = 1^n + 1frac1a_k - a_k - 1 geq frac(n + 1)^2a_n + 1.
$$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, ldots, a_n - a_n - 1$.
answered Mar 21 at 20:42
P. SendenP. Senden
751212
$endgroup$
add a comment |
$begingroup$
We proceed by induction on $n$. For $n = 1$ we have
$$
frac1a_1 - a_0 geq frac1a_1,
$$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
beginalign*
sum_k = 1^n + 1fracn + 2 - ka_k - a_k - 1 &= sum_k = 1^nfracn + 1 - ka_k - a_k - 1 + sum_k = 1^n + 1frac1a_k - a_k - 1 \
& geq sum_k = 1^n frack^2a_k + sum_k = 1^n + 1frac1a_k - a_k - 1,
endalign*
by the induction hypothesis. It is therefore sufficient to prove that
$$
sum_k = 1^n + 1frac1a_k - a_k - 1 geq frac(n + 1)^2a_n + 1.
$$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, ldots, a_n - a_n - 1$.
answered Mar 21 at 20:42
P. SendenP. Senden
751212
$endgroup$
add a comment |
$begingroup$
We proceed by induction on $n$. For $n = 1$ we have
$$
frac1a_1 - a_0 geq frac1a_1,
$$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
beginalign*
sum_k = 1^n + 1fracn + 2 - ka_k - a_k - 1 &= sum_k = 1^nfracn + 1 - ka_k - a_k - 1 + sum_k = 1^n + 1frac1a_k - a_k - 1 \
& geq sum_k = 1^n frack^2a_k + sum_k = 1^n + 1frac1a_k - a_k - 1,
endalign*
by the induction hypothesis. It is therefore sufficient to prove that
$$
sum_k = 1^n + 1frac1a_k - a_k - 1 geq frac(n + 1)^2a_n + 1.
$$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, ldots, a_n - a_n - 1$.
answered Mar 21 at 20:42
P. SendenP. Senden
751212
$endgroup$
We proceed by induction on $n$. For $n = 1$ we have
$$
frac1a_1 - a_0 geq frac1a_1,
$$
which is clearly true, as $a_0 = 0$. Suppose the inequality holds for $n$. Then we have
beginalign*
sum_k = 1^n + 1fracn + 2 - ka_k - a_k - 1 &= sum_k = 1^nfracn + 1 - ka_k - a_k - 1 + sum_k = 1^n + 1frac1a_k - a_k - 1 \
& geq sum_k = 1^n frack^2a_k + sum_k = 1^n + 1frac1a_k - a_k - 1,
endalign*
by the induction hypothesis. It is therefore sufficient to prove that
$$
sum_k = 1^n + 1frac1a_k - a_k - 1 geq frac(n + 1)^2a_n + 1.
$$
This is a straightforward application of the AM-HM inequality to the numbers $a_1 - a_0, ldots, a_n - a_n - 1$.
answered Mar 21 at 20:42
P. SendenP. Senden
751212
answered Mar 21 at 20:42
P. SendenP. Senden
751212
answered Mar 21 at 20:42
P. SendenP. Senden
751212
answered Mar 21 at 20:42
P. SendenP. Senden
751212
751212
add a comment |
add a comment |
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1
$begingroup$
Have you tried induction
$endgroup$
– Jakobian
Mar 21 at 16:58
1
$begingroup$
What have you tried? You'll get a better response on this site if you tell us what you've tried and where you've gotten stuck.
$endgroup$
– jgon
Mar 21 at 17:05
$begingroup$
For $n=1$, $a_0 = -1$, $a_1 = 1$, your inequality is $frac12 geq 1$. It is wrong. Moreover, if one the $a_k$ is $0$, you can't even define the sum on the right. Please precise what numbers you consider.
$endgroup$
– TheSilverDoe
Mar 21 at 17:10
$begingroup$
Why $a_0$ can't be found in RHS?
$endgroup$
– Mostafa Ayaz
Mar 21 at 21:51